It is given that at x = 1, the function x⁴ - 62x² + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.
a = 120
Explanation
This problem requires understanding of maxima and minima on a closed interval. Since x = 1 is given as the point of maximum value on [0, 2], it must be a critical point where the derivative equals zero. The derivative is found, set to zero at x = 1, and solved for a. The second derivative test confirms it's a maximum, and comparing function values at endpoints verifies it's the absolute maximum on the interval.
Solution Steps
Step 1: Let f(x) = x⁴ - 62x² + ax + 9. Since f(x) attains its maximum value at x = 1 on [0, 2], x = 1 must be a critical point.
Step 2: Find the derivative: f'(x) = 4x³ - 124x + a
Step 3: For x = 1 to be a critical point, f'(1) = 0: f'(1) = 4(1)³ - 124(1) + a = 0 → 4 - 124 + a = 0 → a = 120
Step 4: Verify maximum using second derivative: f''(x) = 12x² - 124. At x = 1: f''(1) = 12(1)² - 124 = -112 < 0. Since f''(1) < 0, x = 1 is a point of local maximum.
Step 5: Verify absolute maximum on [0, 2]: f(0) = 9, f(1) = 1 - 62 + 120 + 9 = 68, f(2) = 16 - 248 + 240 + 9 = 17. Since f(1) = 68 is the largest value, x = 1 gives the absolute maximum. Hence, a = 120.